# Sampling Theorem

Joined May 19, 2013
214
Sampling theorem says that if the samples are separated by 1/(2B) seconds then we don't lose information.
My questions:
1) is the sampling done with deltas?
2) if the sampling is done with deltas, then in real life we don't have deltas. Thus, we use the zero-order hold sample process where we lose information. But if we hold the value constant for less than 1/(2B) seconds , would this not mean that we don't lose any kind of information, even in real life?

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#### Kermit2

Joined Feb 5, 2010
4,162
sampling is always a trade off between getting enough resolution in your data and living with the 'missing' parts that can be considered inconsequential if you cannot afford the processing power or manufacturing costs to attain that last .1% or the last .01%.

digitial is never going to be analog. it has to look at a specific instant in time and capture it, then move on to another specific instant each seperated by a discreet amount of time. You will Always be missing information by sampling a data stream.

#### Kermit2

Joined Feb 5, 2010
4,162
it CAN get so close to analog that humans don't seem to give a damn...which is good I think or is it bad.

Hmmm? 1 or 0?

Joined May 19, 2013
214
I know.
But you did not answer my questions.

The Nyquist-Shannon sampling theorem says that a continuous waveform can be completely determined from a discrete waveform if we sample with at least 2B samples per second. Anything lower introduces distortion. And this sampling I think is done with deltas.

My question is. If I hold the value constant for less than 1/2B, would I get a completely determined continuous waveform or not?
Or even if I keep it constant less than 1/2B sec, I would still have distortion but not because of the fact that my sampling rate was somehow higher than 2B, but because of the fact that when I keep it constant, I introduce myself distortion.

Am I saying something right?

#### Kermit2

Joined Feb 5, 2010
4,162
I know

Digital is NEVER going to be analog, but it CAN come so close that it pretty much makes no difference. Although the difference is still there.

Joined May 19, 2013
214
If you know that you did not answer my questions, does this not mean that your own answer has nothing to do with my questions?

I am talking about the impulse sampling and the real sampling. The connection between the perfect world and our world.

You talk only about the real sampling. I know that digital is never going to be analog in the real world. But in theory digital can be analog according to Nyquist-Shannon theorem where we sample with deltas.

So, I am asking about the bridge between the perfect world and our world. Where the bridge breaks. Is this bridge the fact that even though the sample rate is larger than 2B, I still get distortion because I kept the value constant? And I did not take the instantaneously value that the delta would have taken?

I think that I just answered my own question. Thanks, nevertheless.

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#### Kermit2

Joined Feb 5, 2010
4,162
The bridge only breaks when you try to cross it from theory land into reality land.

where is the budget for this project, and how much will people pay for it... etc etc. etc.

you keep asking where the connection between worlds is in theory but keep sidestepping my answer like you don't really believe that it is correct.

Joined May 19, 2013
214
I know that too. But that is how my mind is constructed. It likes to go from ideal to reality in order to understand the reality in a proper way.

#### Kermit2

Joined Feb 5, 2010
4,162

It is a choice of sampling rates. Pick one.

When we got past electron proton neutron we couldn't drive our mathmobile any further down that road either. So we got QUANTUM understanding. A digital sampling of an analog process we humans can't quite get our heads around.

The process is just reversed here in that you want to know when a bunch of 1's and 0's is sufficient to BECOME reality.

It is sufficient to FOOL us, but it will NEVER be anything but ones and zeros

Joined May 19, 2013
214
No, I got it wrong.I did not answer my question above in a correct way.
The correct way is this, I think.

Nyquist-Shannon theorem says that if I have a signal with bandwidth B and if I sample it with a rate equal or greater than 2B then, by associating to each sample a sinc and then adding all those sincs, I obtain the original continuous time waveform.

In the zero-hold process, I might have samples separated by intervales of 1/2B seconds or less, but I dont associate sincs to them and then add them up. I just hold the value constant at each step. Hence, distortion.

And even if I could associate sincs to them, in real life I dont have sincs. They are just mathematical functions extending from -infinite to infinite.

And even if sincs really existed, then there is still the condition that only specific functions are considered within the theorem.

So sinc is really the bridge, not the delta, like I said above.

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#### Kermit2

Joined Feb 5, 2010
4,162
now that you have the right answer, how to represent that DATA brings out the beast of the problem digital vs analog faces.

you are taking a continuous 'stream' of something and trying to make a 'model' of it out of little building blocks called bits. Ones and zeros.
That will NEVER be equivalent to the original stream. It may come so close that nothing different can be detected but it is still just a model built out of data. You can FOOL someone with it, but you will NOT have a true representation of reality.

Even our attempts to capture analog data fail to get the complete picture recorded due to limitations of the equipment we use.

How DO you represent the data contained in your 'syncs'? How does that bring you to the original continuous time waveform without still itself being composed of discrete 'bits' and thereby negating a true reconstruction of the sought after original analog signal?

Joined May 19, 2013
214
In theory, you can reconstruct perfectly the continuous time waveform from discrete samples using sincs. I dont know too much theory, but I suppose that the information that the analog signal (not any arbitrary random signal, but a specific one that has some certain properties) has is carried all into the discrete samples through the sinc function.

In practice, there is no such thing as a sinc function. Thus, information will be lost.
I don't really understand what you are trying to say. I thought everything cleared up nicely.

#### Kermit2

Joined Feb 5, 2010
4,162
yes yes
theory is all tied up nice and neat. with a bow even.

but and it is a BIG butt.

how do you reconcile the claim that you have 're-created' faithfully the analog information while being forced to deal with it as a collection of ones and zeros in whatever processor you create? Is it not still just a MODEL of the analog information? A VERY GOOD MODEL indeed, but still.

Joined May 19, 2013
214
I have another question, though. Samples per bit.
What is this concept?
I thought that, Ok, we have 10 samples, and each sample is on 8 bits. And I want to send these samples.
So I send the first 8 bits of the first sample, then the next 8 bits of the second sample, and so on. And I send them with some bit time.

But what is with this notion of sending multiple samples per bit? It says that the bit time is equal to the samples per bit times the sampling period.

But how do we actually send the samples?

Joined May 19, 2013
214
yes yes
theory is all tied up nice and neat. with a bow even.

but and it is a BIG butt.

how do you reconcile the claim that you have 're-created' faithfully the analog information while being forced to deal with it as a collection of ones and zeros in whatever processor you create? Is it not still just a MODEL of the analog information? A VERY GOOD MODEL indeed, but still.
Is a model, but it is the same as the initial analog signal. Look, I didn't study the theory in detail but counter-intuitive ideas are all over the place in mathematics. For me it is reasonable to think that the discrete samples can contain all the information of an analog signal, under some conditions.
For example, zeno's paradox. In theory, you can never reach that point if you always half the distance because you would need an infinite number of steps. But in practice you reach it.
I am talking about theory not practice.

Joined May 19, 2013
214
Yes. The pdf says:
"The difficulty with the Nyquist-Shannon sampling theorem is that it is based on the notion that the signal to be sampled must be perfectly band limited. This property of the theorem is unfortunate because no real world signal is truly and perfectly band limited. In fact, if a signal were to be perfectly band limited—if it were to have absolutely no energy outside of some finite frequency band—then it must extend infinitely in time."

I know, but I am talking about theory. In theory you can have a perfectly band limited signal.

I see that it also talks about zero-order hold. So is good material. Thanks.

Joined May 19, 2013
214
What is this "samples per bit"?

I know what bits per sample is. But "samples per bits"? And on the internet I can't find anything.
Is the first time I encounter this notion.

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#### Kermit2

Joined Feb 5, 2010
4,162
just going by the info in the picture you posted.

The upper 'bit' is what I would have refered to as sample frequency and the lower bit I would have called resolution, or how many levels are you going to break your sample down into. will you use 8 bits or 16 bits or 20 bits or 24 bits to represent the sample.